((x^3+x^2+x))/((9x^2-25))>= 0

Lösung

\frac{( x ^{3} + x ^{2} + x )}{( 9 x ^{2} - 25 )} \geq 0

- \frac{5}{3} < x \leq 0 or x > \frac{5}{3}

Intervall-Notation

(- \frac{5}{3}, 0] \cup (\frac{5}{3}, \infty)

Dezimale

- 1.66666 \dots < x \leq 0 or x > 1.66666 \dots

Schritte zur Lösung

\frac{x ^{3} + x ^{2} + x}{9 x ^{2} - 25} \geq 0

Faktor

\frac{x ^{3} + x ^{2} + x}{9 x ^{2} - 25} : \frac{x ( x ^{2} + x + 1 )}{( 3 x + 5 ) ( 3 x - 5 )}

\frac{x ( x ^{2} + x + 1 )}{( 3 x + 5 ) ( 3 x - 5 )} \geq 0

Identifiziere die Intervalle

- \frac{5}{3} < x \leq 0 or x > \frac{5}{3}

(5 - x) \cdot (2 x - 6) > 0

2 x^{3} - 3 x^{2} + 7 x - 3 > 0

\frac{( x + 1 )}{( x - 1 ^{2} )} < - 1

\frac{( 3 x - 2 )}{( 9 x + 1 )} < \frac{1}{3}

\frac{( x ^{2} - 2 x + 3 )}{( x + 1 )} \leq 1